The scores on a statewide history exam were normally distributed with $\mu = 78.32$ and $\sigma = 2$. Umaima earned a n $84$ on the exam. Umaima's exam grade was higher than what fraction of test-takers? Use the cumulative z-table provided below. z.00.01.02.03.04.05.06.07.08.09 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974 2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981 2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986
A cumulative z-table shows the probability that a standard normal variable will be less than a certain value (z) In order to use the z-table, we first need to determine the z-score of Umaima's exam grade. Recall that we can calculate her z-score by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}} = \dfrac{84 - {78.32}}{{2}} = 2.84} $ Look up $2.84$ on the z-table. This value, $0.9977$ , represents the portion of the population that scored lower than $84$ on the exam. Umaima scored higher than $99.77\%$ of the test-takers on the history exam.